1
$\begingroup$

I have the following:

If $f:L^2(\mathbb{R})\to H$ is locally lipschitz with $f(0)=0$ and $\langle f(u),u\rangle \leq 0$. Define $$F(u)=\int_{0}^1\langle f(tu),u\rangle dt$$ with $u\in L^1(\mathbb{R})$ and $H \subset L^2(\mathbb{R})$. Show that $F(0)=0 , F(u)\leq 0$ and $f(·)=\nabla F(·)$ on $L^2(\mathbb{R})$.

I need to check the calculations of $f(·)=\nabla F(·)$, the theoretical part not interest me much. I tried derivative $F$ with respect to $u$, but did not get anything.

Any help is appreciated

$\endgroup$
  • 1
    $\begingroup$ What do you mean by $\langle f(u), u\rangle$ when $u$ is only an $L^1$ function? One needs $L^2$ to speak of inner products. $\endgroup$ – nullUser Mar 26 '16 at 23:52
  • 1
    $\begingroup$ you are right, fixed $\endgroup$ – Nightwing Mar 26 '16 at 23:55
  • $\begingroup$ And what is meant by $\nabla F$ when the domain of $F$ is $L^2$ functions? $\endgroup$ – nullUser Mar 26 '16 at 23:59
  • $\begingroup$ Do you have assumptions like $\langle df(x) u, v\rangle = \langle df(x) v, u\rangle$? $\endgroup$ – user99914 Mar 27 '16 at 5:59
  • $\begingroup$ no, only what comes out there.I thought I would write the integral as $ \int_{0}^u f (v) \cdot dv$ making $ v = tu; dv = udt $ then derived. $\endgroup$ – Nightwing Mar 27 '16 at 6:10
1
$\begingroup$

I will prove your claim under the extra assumption that

$$\tag{1} \langle df(x) u, v\rangle = \langle df(x)v, u\rangle,\ \ \ \forall x, u, v\in L^2(\mathbb R).$$

By definition and $(1)$, we have

$$\begin{split} dF(u)(v) &= \int_0^1 \langle df(tu)(tv), u\rangle+ \langle f(tu), v\rangle dt \\ &= \int_0^1 t\langle df(tu)v, u\rangle+ \langle f(tu), v\rangle dt\\ &= \int_0^1 t\langle df(tu)u, v\rangle+ \langle f(tu), v\rangle dt \end{split}$$

Since $$\frac{d}{dt} \langle f(tu), v\rangle = \langle df(tu) u, v\rangle,$$

using integration by part we have

$$\begin{split} \int_0^1 t\langle df(tu)u, v\rangle dt &= t\langle f(tu), v\rangle\bigg|_0^1 - \int_0^1 \langle f(tu), v\rangle dt\\ &= \langle f(u), v\rangle - \int_0^1 \langle f(tu), v\rangle dt. \end{split} $$ Thus $$ dF (u) v =\langle f(u), v\rangle,\ \ \ \forall u, v$$ and so $f(\cdot) = \nabla F (\cdot)$.

On the other hand, the following shows that $(1)$ is almost necessary:

If there is a $C^2$ function $G : L^2(\mathbb R) \to \mathbb R$ so that $\nabla G = f$, then $(1)$ holds.

Proof: Let $x, u, v \in L^2(\mathbb R)$ consider $$A(s, t) = G(x + su + tv)$$ Since $G$ is $C^2$, then so is $A$. Thus partial derivatives commute and $$ \langle df(x) u, v \rangle = \partial_s \partial_t A|_{s=t=0} = \partial_t \partial_s A|_{s=t=0} = \langle df(x) v, u \rangle.$$

Aside: In finite dimensional cases, Your $F$ is a standard way to show that all closed (that is, $(1)$ holds) one form $f$ on $\mathbb R^N$ is exact.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.